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CCSS.Math:

let's solve a few more systems of equations using elimination but in these it won't be Oh kind of a one-step elimination we're going to have to massage the equations a little bit in order to prepare them for elimination so let's say that we have an equation 5x minus 10 y is equal to 15 and we have another equation 3x minus 2y is equal to 3 and I said we want to do this using elimination once again we could use substitution we could graph both of these lines and figure out where they intersect but we're going to use elimination but the first thing you might say hey Sal you know with an elimination you are subtracting the left-hand side of one equation from another or adding the two and then adding the two right-hand sides and I could do that because it's essentially adding the same thing to both sides of the equation but here it's not obvious that that would be of any help if we added these two left-hand sides you would get 8 X minus 12 y that wouldn't eliminate any variables and on the right-hand side you would just be left with a number and if you subtracted that wouldn't eliminate any variable so how is elimination going to help here and the answer is we can multiply both of these equations in such a way that maybe we can get one of these terms to cancel out with one of the others so let's say we want and you could really pick which term you want to cancel out let's say we want to cancel out the Y terms so I'll just rewrite this 5x minus 10 Y here 5x minus 10 y is equal to 15 now is there anything that I can multiply this Green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y so I essentially want to make this negative 2y into a positive 10 Y right because if this was a positive 10 Y it'll cancel out when I add the left-hand sides of this equation so what can I multiply this equation by well if I multiply it by negative 5 negative 5 times negative 2 right here would be positive 10 so let's do that let's multiply this equation times five times negative five so you multiply the left-hand side by negative five and multiply the right-hand side by negative five and what do you get remember we're not chained fundamentally changing the equation we're not changing the information in the equation we're doing the same thing to both sides of it so the left-hand side of the equation becomes negative five times 3x is negative 15 X and the negative 5 times negative 2 y is plus 10y is equal to three times negative five is negative 15 and now we're ready to do our elimination if we add this to the left-hand side of the yellow equation and we add the negative 15 to the right-hand side of the yellow equation we are adding the same thing to both sides of the equation because this is equal to that so let's do that let's do that so 5x minus 15 y we have this little negative sign there we don't want to lose that that's negative 10 X the Y's cancel out negative 10y plus 10y that's 0y that was the whole point behind multiplying this by negative 5 is going to be equal to 15 minus 15 is zero so negative 10 X is equal to zero divide both sides by negative 10 by negative 10 and you get X is equal to zero and now we can substitute back into either of these equations to figure out what Y must be equal to let's substitute into the top equation so we get five times zero minus 10y is equal to 15 or negative 10y is equal to 15 let me write that negative 10y is equal to 15 divide both sides by negative 10 both sides by negative 10 and we are left with Y is equal to 15 over 10 is negative 3 over 2 so if you were to graph it the point of insert intersection would be the point zero rrrow negative three-halves and you can verify that it also satisfies this equation the original equation over here was 3x minus 2y is equal to 3 3 times 0 which is 0 minus 2 times minus 2 times negative 3 halves is this is 0 this is positive 3 right these cancel out these become positive plus positive 3 is equal to 3 so this does indeed satisfy both equations let's do another one of these where we have to multiply and to massage the equations and then we can eliminate one of the variables let's do another one let's say we have let's say we have 5x 5x plus 7y is equal to 15 and we have 700 do another color 7x minus 3y is equal to 5 now once again if you just add it or subtract it both the left-hand sides you're not going to eliminate any variables these aren't in any way kind of have the same coefficient or the negative of their coefficient so let's pick a variable to eliminate let's say we want to eliminate and let's say we want to eliminate the x's this time and you could literally pick on one of the variables or another doesn't matter you could say let's eliminate the Y's first but I'm going to choose to eliminate the x's first and so what I need to do is massage one or both of these equations in a way that these guys have the same coefficients or their coefficients are the negatives of each other so that when I add the left-hand sides they're going to be cut they're going to eliminate each other now there's nothing obvious you know I can multiply this by a fraction to get make it equal to negative 5 or I can multiply this by a fraction to make it equal to negative 7 but even a more fun thing to do is I can try to get both of them to be their least common multiple I could get both of these to 35 and the way I can do it is by multiplying by each other so I can multiply this top equation by 7 let me multiply this top equation by 7 and I'm picking 7 so that this because a35 and I could multiply this bottom equation by negative five by negative 5i negative five and the reason why I'm doing that is so that this becomes a negative 35 or remember my point is I want to eliminate the X's so if I make this a 35 and if I make this a negative 35 then I'm going to be all set I can add the two the left hand and the right hand sides of the equations so this top equation when you multiply by seven it becomes let me scroll up a little bit when you multiply by seven it becomes 35 X + 49 y 49 Y is equal to let's see this is 70 plus 35 is equal to 105 right 15 at 70 plus 35 is equal to 105 that's what the top equation becomes this bottom equation becomes negative 5 times 7 X is negative 35 X negative 5 times negative 3 y is plus 15y all right the negatives cancel out and then 5 this isn't 2 minus 5 this is times negative 5 5 times negative 5 is equal to negative 25 now we can start with this top equation and add the same thing to both sides where that same thing is negative 25 which is also equal to this expression so let's add the left-hand sides and the right-hand sides because we're really adding the same thing to both sides of the equation so the left-hand side the X is canceled out 35 X minus 35 X that was the whole point they cancel out and on the Y's you get 49 y plus 15y that is 64 y 64 y is equal to 105 minus 25 is equal to 80 divide both sides by 64 divide both sides by 60 4 and you get Y is equal to 80 over 64 and let's see if you divide the numerator and denominator by 8 actually if I do 16 16 would be better well let's do 8 first just because we know our 8 times tables so that becomes 10 over 8 and you can divide this by 2 and you get 5 over 4 if you divide it just straight up by 16 you would have gone straight to 5 over 4 so Y is equal to 5 over 4 let's figure out what X is so we can substitute either into one of these equations or into one of the original equations let's substitute into the second of the original equations where we had 7x minus 3y is equal to 5 that was the original version of the second equation that we later transformed into this so we get 7x minus 3 times y times 5 over 4 is equal to 5 or 7x minus 15 over 4 is equal to 5 let's add 15 over 4 oh sorry I didn't do that right this would be 7x minus 3 times 4 oh sorry that was right what am i doing 3 times its 15 over 4 15 over 4 is equal to 5 let's add 15 over 4 to both sides so plus 15 over 4 plus 15 over 4 and what do we get the left-hand side just becomes a 7x these guys cancel out and that's going to be equal to 5 is the same thing as 20 over 4 20 over 4 plus 15 over 4 or we get that let me scroll down a little bit 7x is equal to 35 over 4 we can multiply both sides by 1/7 or we could divide both sides by 7 same thing let's multiply both sides by 1/7 the same thing is dividing by 7 so these cancel out and you're left with X is equal to here if you divide 35 by 7 you get 5 if you divide 7 by 7 you get 1 so X is equal to 5/4 as well so the point of intersection of this of this right here is both x and y are going to be equal to 5/4 so if you looked at it as a graph would be 5/4 comma 5/4 and let's verify that this satisfies the top equation if you take 5 times 5 over 4 plus 7 times 5 over 4 what do you get it should be equal to 15 so this is equal to 25 over 4 plus what is this this is plus 35 35 over 4 which is equal to 60 over 4 which is indeed equal to 15 so it does definitely satisfy that top equation and you can check out this bottom equation for yourself but it should because we actually use this bottom equation to figure out that X is equal to 5/4